following question jumped into my mind.
Let $X$ be a Banach space. Let $C$ be a nonempty subset of $X$. Set
$$A : = \{ x \in X \;|\; \exists \text{ a norm-bounded net } x_{i} \to x \text{ (weakly) } \, \forall i: x_i \in C \}$$
Question 1 : Is $A$ actually the entire weak-closure of C?
Question 2 : Is $A$ weakly closed set?
I'm positive that the answer of 1 is NO, but dont have explicit example. This is because in any infinite dimension space one can construct a net $x_i$ such that $\| x_i \| \to \infty$ but $x_i \to 0~(weakly)$.
Question 2 is harder, I tried to prove it, I had to go through some diagonal process regarding for index sets, which was kinda impossible !! Now I changed my mind, and start to find a counter example instead of proving it.
Any help would be greatly appreciated.
The answer to both questions is negative. Here are some counterexamples in $\ell^2$.
For $$C := \{ e_n + n \, e_m \mathrel\mid n,m \in \mathbb N,\; n < m \}$$ one hase $$A = C \cup \{ e_n \mathrel\mid n \in \mathbb N \}$$ but the weak closure of $C$ contains $0$. The set $A$ is also not weakly closed.
For another example, you can take $$ C:= \{ \sqrt{n} \, e_n \mathrel\mid n \in \mathbb N\}.$$ Then, $A = C$, but again, the weak closure of $C$ contains $0$.
Conclusion: nets are weird.